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Reversible Reactions and Chemical Equilibrium. University of Lincoln presentation. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License. Outline. Reversible reactions Chemical Equilibrium Le Chatelier’s Principle

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Reversible Reactions and Chemical Equilibrium

University of Lincoln presentation

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Outline

Reversible reactions

Chemical Equilibrium

Le Chatelier’s Principle

Equilibrium constants

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Reversible Reactions

BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)

CH3CO2H + CH3CH2OH ↔ CH3CO2CH2CH3 + H2O

Cr2O72-(aq) + 2OH-(aq) ↔ 2CrO42-(aq) + H2O(l)

CH3CO2H(aq) + H2O(l) ↔ CH3CO2-(aq) + H3O+(aq)

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Chemical Equilibrium

Reactions not 100% complete

Products and Reactants exist together

A dynamic equilibrium

Position of equilibrium ???

Can the position of equilibrium be changed?

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Le Chatelier’s Principle

When an external change is made to a system in equilibrium, the system will respond to oppose the change

External Changes

Concentration

Pressure (gases)

Temperature

Link to external video

Link to external video

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Concentration

1. BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)

2. Cr2O72-(aq) + 2OH-(aq) ↔ 2CrO42-(aq) + H2O(l)

How does reaction 1 respond to addition of hydrochloric acid?

How does reaction 2 respond to addition of alkali?

How does reaction 2 respond to addition of acid?

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Pressure

N2(g) + 3H2(g) ↔ 2NH3(g)

CO(g) + 2H2(g) ↔ CH3OH(g)

2NO2(g) ↔ 2NO(g) + O2(g)

PCl5(g) ↔ PCl3(g) + Cl2(g)

H2(g) + I2(g) ↔ 2HI(g)

CO(g) + H2O(g) ↔ CO2(g) + H2(g)

How do the above equilibria respond to:

An increase in pressure

A decrease in pressure

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Temperature

N2(g) + 3H2(g) ↔ 2NH3(g) rH = -92.2 kJ mol-1

H2(g) + I2(g) ↔ 2HI(g)rH = -9.4 kJ mol-1

CO(g) + H2O(g) ↔ CO2(g) + H2(g) rH = -41.2 kJ mol-1

PCl5(g) ↔ PCl3(g) + Cl2(g) rH = 87.9 kJ mol-1

How do the above respond to an

Increase in temperature

Decrease in temperature

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Equilibrium constantsa measure of equilibrium position

aA + bB ↔cC + dD

BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)

Write the expressions for Kc for the reactions given in previous slides

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Calculating Equilibrium Constants

HNO2(aq) ↔ H+(aq) + NO2-(aq)

The table shows the equilibrium molar concentrations for three solutions of nitrous acid in water at 25 oC

Calculate the equilibrium constant for this reaction at 25oC

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Solution A

Units of Kc

Now try for solutions B and C

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Acids and Bases

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Outline

Definitions

Weak Acids

Dissociation Constants

Weak Bases

Drugs

pH

Buffers

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Acids and Bases

Several definitions available - most common is Bronsted and Lowry

Acid is a proton donor

HCl is able to transfer H+

Base is a proton acceptor

NH3 is able to accept H+ and become NH4+

Aqueous solutions

Proton species is H3O+ (hydroxonium ion)

HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)

HCl(aq)  H+(aq) + Cl-(aq)

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Strong Acids

Strong acids are fully dissociated

HCl (aq) + H2O(l) H3O+ (aq) + Cl- (aq)

all dissolved HCl molecules are ionised

1 mol dm-3 HCl(aq) there are:

Approx 1 mol dm-3 H3O+ (aq)

Approx 1 mol dm-3 Cl- (aq)

DO NOT confuse ‘strong’ and ‘concentrated’

1 x 10-4 mol dm-3 HCl (aq) is a dilute solution of a strong acid

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Other strong acids

HNO3 (nitric)

H2SO4 (sulfuric)

HClO4 (perchloric)

Write equations showing the dissociation of the above acids

Which are monoprotic?

Are any diprotic?

Chemical equilibrium – K very large

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Weak Acids

Acids that dissociate in a reversible reaction (e.g. CH3COOH; ethanoic (acetic) acid)

CH3COOH (aq) + H2O(l) ↔ H3O+ (aq) + CH3COO- (aq)

Solution of CH3COOH (aq) contains:

CH3COOH (aq)

H3O+ (aq)

CH3COO- (aq)

CH3COOH is partially dissociated

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How weak is a weak acid?

0.1 mol dm-3 HCl is dissociated 91.4%

[H3O+] = 0.091 mol dm-3pH=1.04

0.1 mol dm-3 CH3COOH is dissociated 1.34%

[H3O+] = 0.0013 mol dm-3pH=2.87

Extent given by K

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Weak Acids

HA(aq) + H2O(l)↔ H3O+(aq) + A-(aq)

HA Bronsted acid

H2O Bronsted base

H3O+ Bronsted acid

A- Bronsted base

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Acid dissociation constant (Ka)

The higher the Ka value:

greater degree of ionisation

stronger the acid

Data tables

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Ka Values

HCO2H1.8 x 10-4 mol dm-3

CH3CO2H1.7 x 10-5 mol dm-3

Are these weak or strong acids?

Which is the stronger acid?

HCO2H3.75

CH3CO2H4.77

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pKa values (data tables)

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pKa Values

Controlling the ionisation of weak acids

pH = pKa then [HA] = [A-]

pH > pKa then [A-] > [HA]

pH < pKa then [HA] > [A-]

CH3COOH (aq) + H2O(l) ↔ H3O+ (aq) + CH3COO- (aq)

CH3COOH: CH3COO- at pH = 4.77 ?

CH3COOH: CH3COO- at pH = 3 ?

CH3COOH: CH3COO- at pH = 7 ?

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Henderson-Hasselbach

For weak acids

Use the equation with the example in the previous slide. Do you come to the same conclusion regarding the ratio of un-ionised to ionised acid molecules?

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Weak Bases

B(aq) + H2O(l) BH+(aq) + OH-(aq)

CH3NH2(aq) +H2O(l) ↔ CH3NH3+(aq) + OH-(aq)

pKa = 10.66 (of conjugate acid) [B]=[BH+]

pH = 10.66

pH =8 what happens to CH3NH3+(aq): CH3NH2(aq)

pH =13 ?

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Henderson-Hasselbach

For weak bases

Use the equation with the example in the previous slide. Do you come to the same conclusion regarding the ratio of un-ionised to ionised acid molecules?

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Acidic drugs

2-[4-(2-methylpropyl)phenyl]propanoic acid

ibuprofen

How does this molecule ionise?

pKa=4.5

pH =3 (stomach pH)?

pH=6 (intestine)?

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Basic drugs

amphetamine (C6H5CH2CH(NH2)CH3)

Write an equation for the reaction of amphetamine with water.

The pKa of the conjugate acid is 9.8. What will happen to the ratio of ionised to unionised amphetamine at:

pH 7

pH 12

Why might this be important?

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Water

Can dissociate:

H2O(l) ↔ H+(aq) + OH-(aq)

2H2O(l) ↔ H3O+(aq) + OH-(aq)

H2O is amphoteric

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Water

Kw = [H3O+][OH-]= 1 x 10-14 mol2 dm-6

Kw the ionic product of water

In pure water what is [H3O+] and [OH-] ?

Kw is a very smallconstant

water is only very partially ionised

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pH

pH is defined as:

pH = -log10[H3O+]

pH is a measure of the H3O+ concentration in solution and can vary from 1 to 14

pH=7 – neutral [H3O+] = [OH-]

= 1 x 10-7 mol dm-3 at 25 oC

pH<7 – acidic [H3O+] >[OH-]

pH>7 - alkaline/basic [H3O+] <[OH-]

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pH-examples

0.1M HNO3

0.1M CH3COOH

What is the pH?

pH is dependent on the ionisation of the acid

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pH-examples

What about alkaline solutions?

E.g. 0.1M NaOH solution

Will also depend on degree of ionisation

use equation: [H+] x [OH-] = 10-14

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Buffers

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Buffers

A buffer solution resists pH changes on addition of small amounts of acid or base (alkali) to a system.

Very important

e.g. blood has a pH of 7.4. If it varies by ± 0.4, death can occur

Buffer solutions rely upon the effects of a weak acid or base and the salt of that acid or base

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Buffers

Ethanoic acid (a weak acid) and sodium ethanoate (salt)

CH3COOH  CH3COO- + H+(1)

CH3COONa  CH3COO- + Na+ (2)

(1)-partially ionised

(2)-fully ionised

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Buffers

Henderson-Hasselbach equation

Acidic buffers

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Making a buffer solution

Choose a weak acid with a pKa close to the required pH of the buffer.

Choose an appropriate salt of the weak acid

Determine [salt]/[acid] ratio needed to give correct pH

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An acidic buffer: Ethanoic acid and sodium ethanoate

  • What would be the pH of an ethanoate buffer with equal acid and sodium ethanoate concentrations?

What is the [salt] if the acid is 0.1 mol dm-3 to give buffer solutions of

pH = 5

pH = 4

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An alkaline buffer:ammonia solution and ammonium chloride

Note the base/salt ratio

What is the pH of a buffer with base:salt ratio = 1?

Calculate the base:salt ratios for pH 8.5 and pH 10.5

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Acknowledgements

  • JISC

  • HEA

  • Centre for Educational Research and Development

  • School of natural and applied sciences

  • School of Journalism

  • SirenFM

  • http://tango.freedesktop.org

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